Timeline for A question regarding Corollary 4.12 in Mumford's "Analytic construction of deg. Ab. Var."
Current License: CC BY-SA 4.0
18 events
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| Mar 16, 2022 at 16:10 | comment | added | Laurent Moret-Bailly | If $[p]:G\to G$ is not finite, I suggest you look at the image $H$ of $[p^n]$ for large $n$. It is ultimately independent of $n$, connected, $[p]:H\to H$ is surjective, and if $X$ is as before, then $p^nX=H$. (Here I assume that $G/G^0$ is a $p$-group; we can always reduce to that case). If someone has a reference or a simpler argument, I'll buy it. | |
| Mar 16, 2022 at 16:01 | comment | added | Laurent Moret-Bailly | @TheThinWhistler: For a commutative algebraic group $G$ over an algebraically closed field $k$, and any prime $p$, I claim that if the $p$-torsion of $G$ is $p$-divisible then $G/G^0$ has order prime to $p$. This is easy if $[p]:G\to G$ is finite (in particular, if $p\neq\mathrm{char}(k)$): if $X$ is a connected component of maximal $p$-power order $p^m$, then $p^mX=G^0$, so $X$ contains an element of order $p^m$ which is not divisible by $p$ if $m>0$. | |
| Mar 16, 2022 at 9:42 | history | edited | The Thin Whistler | CC BY-SA 4.0 |
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| Mar 16, 2022 at 9:41 | comment | added | The Thin Whistler | @LaurentMoret-Bailly why is it so clear that $p$-divisibility of the torsion implies connectedness? Excuse me if I am stupid. | |
| Mar 16, 2022 at 7:11 | history | edited | The Thin Whistler | CC BY-SA 4.0 |
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| Mar 14, 2022 at 20:44 | comment | added | Jason Starr | Okay, now I understand the issue. I still suggest reduction mod $p$. For every characteristic $0$ fraction field $K$ of a finite type $\mathbb{Z}$-algebra $A$ that is an integral domain, for every finite type group $K$-scheme $G_K$, and for every nonconstant morphism of group $K$-schemes from $\mathbb{A}^1_K$ to $G_K$, after inverting a nonzero element of the integral domain, all of this extends over $\text{Spec}\ A$. If we know that the reduction modulo all but finitely many primes $p$ in $\mathbb{Z}$ has $p$-divisible group of $p$-torsion points, this gives a contradiction. | |
| Mar 14, 2022 at 13:29 | comment | added | Laurent Moret-Bailly | @JasonStarr: If we know in general that the semi-abelian locus is open, then we can invoke this directly since the reduction mod $I$ is a torus. Even the computation of torsion is not needed. | |
| Mar 14, 2022 at 11:32 | comment | added | Jason Starr | Now I see the statement. I am going to repeat my earlier suggestion. Mumford does not require that $S$ is a DVR. We can write our group scheme as the base change of a group scheme over a finitely generated $\mathbb{Z}$-algebra, and then complete at an appropriate prime whose residue field has positive characteristic. Then the $p$-divisibility does imply semi-Abelian reduction. Since the semi-Abelian locus in the base scheme is open, that gives a way to prove semi-Abelian reduction over the original scheme. | |
| Mar 14, 2022 at 8:59 | comment | added | Laurent Moret-Bailly | @JasonStarr On top of p. 24, Corollary (4.12). Clearly $p$-divisibility for all $p$ implies connectedness, but for the unipotent radical something seems to be missing. Perhaps the fact that the reduction mod $I$ is a torus is enough, but for non-affine groups I have trouble finding a reference. | |
| Mar 14, 2022 at 1:07 | comment | added | Jason Starr | I see nowhere in Mumford’s article where he uses $p$-divisibility. On what page do you see any claim that uses $p$-divisibility? | |
| Mar 13, 2022 at 23:38 | comment | added | The Thin Whistler | I am not talking about smoothness! I am talking about connectedness and no unipotent radical. | |
| Mar 13, 2022 at 22:54 | comment | added | Jason Starr | I just quickly read Mumford's article. Nowhere in the article does he claim to deduce smoothness of the model from $p$-divisibility. In the introduction he compares his construction to Tate's construction using $p$-adic uniformization. But Mumford does not use that in his construction. | |
| Mar 13, 2022 at 22:30 | comment | added | The Thin Whistler | That doesn't help at all. | |
| Mar 13, 2022 at 20:28 | comment | added | Jason Starr | I do not have access to Mumford's paper. What I can say is that $p$-divisibility rules out certain kinds of bad reduction. Without looking at Mumford's paper at all, one possibility is that Mumford uses the technique (also used elsewhere to prove theorems about Abelian varieties) of showing that the family of Abelian varieties is the base change of a family with similar properties defined over a finite type scheme over $\text{Spec}\ \mathbb{Z}$. If this family also satisfies $p$-divisibility then you are set: just reduce modulo maximal ideals having finite residue fields. | |
| Mar 13, 2022 at 17:28 | comment | added | The Thin Whistler | I mean he shows that $G_s$ is of char $0$. What do I gain from this? | |
| Mar 13, 2022 at 16:59 | comment | added | The Thin Whistler | @JasonStarr So what? Excuse me if I am stupid. | |
| Mar 13, 2022 at 16:44 | comment | added | Jason Starr | Multiplication by $p$ on an additive group / vector group in characteristic $p$ is a zero map. So the additive group is not $p$-divisible in characteristic $p$. | |
| Mar 13, 2022 at 16:12 | history | asked | The Thin Whistler | CC BY-SA 4.0 |